Here are small parts of two **Serpentine Lattice Labyrinth** tessellations.

Though these two patterns look very different and the arrays of superlattice points (which we can take to be marked by the black squares) and symmetry axes are at different inclinations, the supertiles of each tessellation are of the same area – 25 squares.

This equality arises simply because 7² + 1² = 5² + 5² = 50. two supertiles each of area 25 and oriented at right angles to each other constitute the fundamental domain of the tessellation.

I am yet to discover the term for pairs of number pairs such as (7,1) and (5,5) so for the time being will refer to them as ** Pythagorean pairs of pairs**. This is because two right angled triangles of integral lengths 7 and 1 and 5 and 5 respectively share a hypotenuse of equal, though not integral length. (7,1) and (5,5) are the lowest possible such pair of pairs, though if you allow one of the numbers to be zero, the Pythagorean triple represented by (0,5) , (3,4) could be said to be a

*degenerate*or

*special*case of a pair of pairs.

Here a list of all the Pythagorean pairs of pairs (including the triples) for sums of squares up to 1000. I’ve put the lowest numbers first to show how their magnitude build up. Several initial-omitted cases were added on 16th. June. 2015, hence the empty half-rows.

NP stands for *not primitive*, a a multiple of a lower order primitive case. For instance (0,10) , (6,8) is (0,5) , (3,4) multiplied by 2. For the meaning of “Generating pairs” see below.

Note the cases where a particular sum of squares can be arrived at not just in two but in three ways, for instance the sum of squares of (2,29), (13,26) and (22,19) all equal 845.** So far all such triplet sums of squares have 5 as a factor – but then, so do the first eight pair-of-pairs cases in the table, and that didn’t last! **(See the end of this post.)

The earlier post *Eine kleine Mathmusik *explains how we can multiply two number pairs (a,b) and (c,d) together and how this is homologous to making a tessellation of tessellations with these number pairs as separation parameters – a *product tessellation* which repeats according to separation parameters (x,y) where x = ac-bd and y = ad+bc.

here again is the geometry of the example I gave.

If we reverse the order of one number pair, we find that (2,1) x (3,1) = (5,5). Hey presto, we have generated one of the Pythagorean pairs of pairs.

Here is the equivalent tessellation of tessellations.

Instead of realising the product tessellations as a pattern with (3,1) or (1,3) supertile-shapes each made up of the (2,1) crosses, we can reverse the situation and make a pattern of (2,1) cross-shapes each made up of (3,1) or (1,3) supertiles. here are the Pythagorean Pair of possibilities realised in this way.

So now we have no less than six tessellations, each with supertile area 25 and fundamental domain 50, corresponding to the separation parameter Pythagorean pair of pairs (1,7) and (5,5) and we have seen how this pair of pairs is generated by multiplying the smaller number pairs (1,2) and (1,3).

Similarly **EVERY** Pythagorean pair of pairs can be generated by multiplying together two smaller number pairs, hence the *Generating pairs *columns. In each case:

A *generating pair of pairs***(p,q)** and **(r,s)** will produce a Pythagorean pair of pairs: **(a,b) = (pr-qs, ps+qr) , (c,d) = (ps-qr, pr+qs)** so long as none of p,q,r,s are zero and p≠q, r≠s. The sum of squares (a² + b²) = (c² + d²) = ( p² + q²) x ( r² + s²). If one of the parameters is zero or p=q or r=s, then just one number pair and product tessellation is generated. (p,q) x (r,0) = (pr,qr) and (p,q)x(0,r) = (-qr,pr) which generates the same superlattice and tessellation as (pr,qr).

** Note that altering the order and one of the signs of the separation parameter number pair leaves the tessellation unchanged. In general: (x,y) ≡ (-y,x) ≡ (-x,-y) ≡ (y,-x). **One has frequently to make one of these substitutions to keep the parameters positive. Simply swapping the numbers without changing a sign generates the mirror image of the corresponding tessellation, which I often rather lazily take to be the same tessellation, just “flipped over”.

**HOWEVER**, if one parameter is zero and the other three constitute a **Pythagorean triple**, for instance (3,4) and (5,0) then, replacing then hypotenuse number by the other two side lengths we can generate not just a Pythagorean pair of pairs but one of the ** triplets** (not to be c0nfused with a

*triple*)that you can see strewn rather thinly across the above table. For instance:

**(5,0) x (3,4) = (15,20) (4,3) x (4,3) = (7,24) (4,3) x (3,4) = (0,25) ** and we confirm that **15² + 20² = 7² + 24² = 0² + 25² = 625**.

Another way of looking at this triplet is to see that it is a pair of Pythagorean triples sharing the same hypotenuse: 15,20,25 and 7,24,25. The first of these triples is not *primitive*, if we divided by 5 we get 3,4,5. If we divide the second triple by 5 we get 7/5, 24/5, 5 which gets us into the subject of **right-angled triangles with rational non-inegral sides**, which must indeed be a* subject, *but it one about which I know nothing.

How do we generate the triplet of pairs **(5,20), (8,19), (13,16)** whose squares sum to **425**? Well, 425 can be factorised to 17 x 25, corresponding to number pairs (4,1) and (5,0). If we multiply **(4,1) x (5,0)** we get (4×5 – 1×0, 4×0 + 1×5) **= (20,5)**, which I ordered (5,20) in the table To get the other pairs in the table we nore that (5,0) has the same square as (4,3) and that ** (4,1) x (4,3) = (13,16) ** and **(4,1) x (3,4) = (8,19)** **QED**………**BUT**… what about factorising to 85 x 5, corresponding to number pairs (9,2) and (2,1)? Well, (9,2) x (2,1) = (16,13) and (9,2) x (1,2) = (5,20), so we’ve generated nothing new.

The triplet of number pairs whose parameters sum to **725** can be generated as products of **(5,2)** and **(5,0)** manipulated in a similar way as for 425. The triplet for **845**, which factorises to 13² x 5, is generated by **(12,5)**, **(13,0)** and **(2,1)**, that for **850**, which factorises to 17 x 5³, by **(4,1) x (10,5)** and **(5,0) or (4,3) x (9,2)**.

All the above triples are generated by number pairs including (5,0) or (13,0), which can be represented by the other members of Pythagorean triples, namely (4,3) and (12,5). All the triplet (or still more fecund) cases obviously require THREE number pairs to generate them and this can only arise when two of them are made up from a Pythagorean triple.

It’s interesting to investigate what happens if the factors of a sum of squares include TWO Pythagorean triples, say both the pairs (5,0) AND (13,0). If these are the only factors, the number is **5² x 13² =3125**. The number pairs we have available as generating pairs are (5,0), (13,0), (4,3) and (12,5), all of which can of course be reversed in the usual way. The following are the distinct products generated, showing one way of generating each.

**(13,0) x (5,0) = (65,0) (13,0) x (4,3) = (52,39) (12,5) x (5,0) = (60,25)**

**(12,5) x (4,3) = (33,56) (12,5) x (3,4) = (16,63) Each a² + b² = 3125
**

We have jumped from a Pythagorean triplet of pairs with the same sum of squares straight to a Pyuthagorean quintuplet of pairs. (NOT to be confused with another annoying use of *Pythagorean Triples* to mean (a,b,c)’s where a² + b² + c² = d², specifing a three-dimensional equivalent of the two-dimensional right-angled triangle with integer sides) Once again, five is a factor of the sum of squares, but it could hardly not be – I built it in from the start.

How about a number which has **13²** (**13** being the next integral hypotenuse up) but not **5²** as a factor; that should generate at least a triplet of pairs. The lowest number which forms a triplet without 5 as a factor may be **13²x17 =2873**, the number pairs being **(8,53), (13,52) and (32,43)**, generated by multiplying the number pairs **(13,0), (12,5) and (4,1)** and their reversals.

I’ve only added a few of the generating number pairs to the pairs-of-pairs table. In order to show how they trend, I’ve again put the smallest numbers first in the brackets, so multiplying them will give negative answers, but remember that the tessellations (a,b), (-b,a), (b,-a) and (-a,-b) are identical, while( a,-b), (b,a), (-a,b) and (-b,-a) are simply their mirror image, so convertion into positive separation parameter pairs is straightforward.

*Here are some links to the inexpensive how-to-do-it ***Lattice Labyrinths*** workbook which is available from the publisher or you-know-who , or from a good independent bookshop or via Google.*

Reblogged this on Herminio López.

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Belated thanks, DM

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